Difference between revisions of "2006 AMC 10B Problems/Problem 17"

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Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?  
 
Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?  
  
<math> \mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{5}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math>
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<math> \textbf{(A) } \frac{1}{10}\qquad \textbf{(B) } \frac{1}{6}\qquad \textbf{(C) } \frac{1}{5}\qquad \textbf{(D) } \frac{1}{3}\qquad \textbf{(E) } \frac{1}{2} </math>
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== Video Solution by OmegaLearn ==
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https://youtu.be/5UojVH4Cqqs?t=1160
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~ pi_is_3.14
  
 
== Solution ==
 
== Solution ==
Since there is the same amounts of all the balls in Alice's bag and Bob's bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob's bag doesn't matter. [[Without loss of generality]], let the ball Alice puts in Bob's bag be red.  
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Since there are the same amount of total balls in Alice's bag as in Bob's bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob's bag doesn't matter. [[Without loss of generality]], let the ball Alice puts in Bob's bag be red.  
  
For both bags to have the same contents, Bob must select one of the 2 red balls out of the 6 balls in his bag.  
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For both bags to have the same contents, Bob must select one of the <math>2</math> red balls out of the <math>6</math> balls in his bag.  
  
So the desired probability is <math> \frac{2}{6} = \frac{1}{3} \Rightarrow D </math>
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So the desired probability is <math> \frac{2}{6} = \boxed{\textbf{(D) }\frac{1}{3}}</math>.
  
 
== See Also ==
 
== See Also ==
*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=16|num-a=18}}
 
 
*[[2006 AMC 10B Problems/Problem 16|Previous Problem]]
 
  
*[[2006 AMC 10B Problems/Problem 18|Next Problem]]
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 03:32, 4 November 2022

Problem

Bob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?

$\textbf{(A) } \frac{1}{10}\qquad \textbf{(B) } \frac{1}{6}\qquad \textbf{(C) } \frac{1}{5}\qquad \textbf{(D) } \frac{1}{3}\qquad \textbf{(E) } \frac{1}{2}$

Video Solution by OmegaLearn

https://youtu.be/5UojVH4Cqqs?t=1160

~ pi_is_3.14

Solution

Since there are the same amount of total balls in Alice's bag as in Bob's bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob's bag doesn't matter. Without loss of generality, let the ball Alice puts in Bob's bag be red.

For both bags to have the same contents, Bob must select one of the $2$ red balls out of the $6$ balls in his bag.

So the desired probability is $\frac{2}{6} = \boxed{\textbf{(D) }\frac{1}{3}}$.

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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